Question

Please help me!?

Answer 1

can you see the whole problem?

Answer 2

multiply by the reciprocal first then factor everything you can

Answer 3

I am having trouble factoring everything except the 15t-10

Answer 4

\[4t^2-1=(2t)^2-(1)^2=(2t-1)(2t+1)\]

Answer 5

difference of squares on that one

Answer 6

oh okay! i got confused with the 1 for that.. now how would we do 3t^2+5t+2 ?

Answer 7

do 3*2 then get the factors of 6

Answer 8

SHOULD be shown as :-

(2/3) t - ( 2 - 3 t ) < 5 t + 2 ( 1 - t )

2 t - 6 + 9 t < 15 t + 6 - 6 t

2 t < 12

t < 6

Answer 9

so (3t+3) (3t-2) ?

Answer 10

@abayomi12 does not look like we are solving for t

Answer 11

we are just finding the factors!

Answer 12

im not sure kakrazz i dont have any paper with me

Answer 13

but 6 is the answer

Answer 14

wrong question

Answer 15

lol

Answer 16

(3t+3) (3t-2)=9t^2+3t-6

Answer 17

when i find the factors of 6, which are 3 and 2 that adds up to 5, do i keep the 3t or do i factor out that 3?

Answer 18

to just use t? or,

Answer 19

use t

Answer 20

now im confused, because if i just use t, then if i foiled it - it wuold be t^2, not 3t

Answer 21

3t^2+5t+2
3t^2+3t+2t+2

Answer 22

OH OH OH FACTOR BY GROUPING

Answer 23

thats one way of doing it

Answer 24

look at the middle term, see if you can group them into factors of the first and 2nd, then see if its factorable

Answer 25

yes! Ifound that the factors of that equation are (3t+2) (t+1)

Answer 26

now 4t^2-4t+1

Answer 27

\[\frac{4t^2-1}{15t-10}\div \frac{3t^2+5t+2}{4t^2-4t+1}\]

Answer 28

4t^2-4t+1- looks square

Answer 29

i found that teh factors of 4t^2-4t+1 are (2t-1) and (2t+1)

Answer 30

i never said that they were a difference of squares, i just said i was square

(2t-1)(2t+1)=4t^2-1

Answer 31

Oh i know, im tking about 4t^2-4t+1

Answer 32

just looking at the signs, we know the answer is in the form of (x-a)(x-b)
i said it was square so
(x-a)^2